A 2.5 mm -diameter sphere is charged to -4.4 nC . An electron fired directly at

Question

A 2.5 mm -diameter sphere is charged to -4.4 nC . An electron fired directly at the sphere from far away comes to within 0.30 mm of the surface of the target before being reflected.

Part A:

What was the electron's initial speed?

Express your answer using two significant figures. (in m/s)

Part B:

At what distance from the surface of the sphere is the electron's speed half of its initial value?

Express your answer using two significant figures. (in m)

Part C:

What is the acceleration of the electron at its turning point?

Express your answer using two significant figures (in m/s^2)

Explanation / Answer

Here,

charge , Q = -4.4 C

radius of sphere , r = 2.5/2 mm

r = 1.25 mm

part A)

let the initial speed of electron is u

Using conservation of energy

0.5 * m * u^2 = k * q * e/d

0.5 * 9.109 *10^-31 * u^2 = 9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/(0.00125 + 0.3 *10^-3)

solving for u

u = 9.48 *10^7 m/s

the electron's initial speed is 9.48 *10^7 m/s

part b)

for half the initial speed

u = 9.48 *10^7/2 m/s

u = 4.74 *10^7 m/s

Using conservation of energy

0.5 * 9.109 *10^-31 * ( 4.74 *10^7 )^2 = 9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/(0.00125 +d)

d = 0.00495

d = 4.95 mm

distance from the surface of the sphere is the electron's speed half of its initial value is 4.95 mm

part c)

let the accelertion is a

Using second law of motion

9.1 *10^-31 * a = 9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/(0.00125 + 0.3 *10^-3)^2

solving for a

a = 2.901 *10^18 m/s^2

the acceleration at the turn arround point is 2.901 *10^18 m/s^2

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